Question

A cylinder contains \(10 \mathrm{~kg}\) of gas at pressure of \(10^{\prime} \mathrm{N} / \mathrm{m}^{2}\). The quantity of gas taken out of the cylinder, if final pressure is \(2.5 \times 10^{6} \mathrm{Nm}^{-2}\). will be (temperature of gas is constant) (A) \(5.2 \mathrm{~kg}\) (B) \(3.7 \mathrm{~kg}\) (C) \(7.5 \mathrm{~kg}\) (D) \(1 \mathrm{~kg}\)

Short answer

The short answer is: None of the given options are correct. The mass of the gas removed from the cylinder should be \(9.6\, kg\).

Step-by-step solution

Write the given values

We have been given the following information: Initial mass of gas, \(m_i = 10\, kg\) Initial pressure, \(P_i = 10^5\, N/m^2\) Final pressure of gas, \(P_f = 2.5 \times 10^6\, N/m^2\)

Apply ideal gas law

Since the temperature is constant, the ratio of the initial and final pressure and volume remains the same. So, we can write the following equation: \(\frac{P_i V_i}{P_f V_f} = \frac{n_i}{n_f}\) Where \(V_i\) and \(V_f\) are the initial and final volume, and \(n_i\) and \(n_f\) are the initial and final number of moles of gas, respectively.

Solve for the initial and final volume ratio

Using the equation from step 2, we can solve for the ratio of initial and final volume: \(\frac{V_i}{V_f} = \frac{P_f}{P_i} = \frac{2.5 \times 10^6}{10^5} = 25\)

Solve for the mass of gas removed

Since mass and volume are proportional (mass = density * volume), the ratio of the initial and final mass is the same as the ratio of initial and final volume, which we found in step 3: \(\frac{m_i}{m_f} = \frac{V_i}{V_f} = 25\) Now, we can solve for the final mass, \(m_f\): \(m_f = \frac{m_i}{25} = \frac{10\, kg}{25} = 0.4\, kg\) Finally, we can find the mass of the gas removed, \(m_r\), by subtracting the final mass from the initial mass: \(m_r = m_i - m_f = 10\, kg - 0.4\, kg = 9.6\, kg\) However, this result is not available in the given options. It is possible that there was a misprint in the question or the answer choices. Based on the method we have used, the answer should be \(9.6\, kg\).

Key concepts

Pressure and Volume Relationship

According to the Ideal Gas Law, for any gas held at a constant temperature, there is a relationship between the pressure and the volume it occupies. Specifically, this relationship is characterized by the equation: \( PV = nRT \). At a consistent temperature (often noted as being an isothermal process), if the number of moles \( n \) and the gas constant \( R \) are stable, the product of pressure \( P \) and volume \( V \) remains constant.
This leads to the equation: \( \frac{P_i}{P_f} = \frac{V_f}{V_i} \), where \( P_i \) and \( P_f \) are the initial and final pressures, respectively, and \( V_i \) and \( V_f \) are the initial and final volumes.
In simpler terms, if the pressure of a gas increases while its temperature remains constant, its volume must decrease proportionally to maintain the same product \( PV \). This experiment showcases how changing pressures can significantly affect the gas's volume, showing the inverse relationship between these two quantities.

Constant Temperature Process

An isothermal process is defined by constant temperature throughout the entire process. Such a process ensures that any change in the pressure or volume of the gas does not alter its temperature.
In the context of the problem, the constant temperature means that the internal energy of the gas does not change. The change in entropy is completely balanced by the heat exchange with the environment.

• This scenario makes calculations easier since the ideal gas law can be simplified under an isothermal condition.
• Mathematically, it allows us to equate initial and final states with integrity, relying solely on the pressure and volume relationships.

This attribute is what simplifies many calculations in thermodynamics, turning complex processes into more understandable scenarios when temperature remains unchanged.

Gas Removal Calculation

To calculate the quantity of gas removed, we use the solved relationship between initial and final volumes. Since the temperature is stable, the equation \( \frac{V_i}{V_f} = \frac{m_i}{m_f} \) holds true, where \( m_i \) and \( m_f \) represent the initial and final mass of the gas.
The given values allowed us to express this relationship in terms of pressures: thus \( \frac{V_i}{V_f} = \frac{P_f}{P_i} = 25 \). It follows that the mass relationship is \( \frac{m_i}{m_f} = 25 \).
With an initial mass \( m_i \) of 10 kg, the final mass \( m_f \) can be calculated as \( m_f = \frac{10}{25} = 0.4 \) kg. Therefore, the amount of gas removed is computed by subtracting the final mass from the initial mass:

• Initial mass: 10 kg
• Final mass: 0.4 kg
• Gas removed: 9.6 kg

This step-by-step process intricately shows how proportional changes in pressures and volumes directly correlate to the mass of gas left or removed when the temperature remains constant.